0,2

If an integer N is square-free and has n+2 distinct prime factors then a(n) is the number of product signs needed to write the factorizations of N, so a(n)=A076277(N) - Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 17 2002.

Representation as an infinite series, in Maple notation : a(n-1)=sum(k^n*k*(k-2)/k!, k=1..infinity)/exp(1), n=1, 2... . This is a Dobinski-type summation formula. - Karol A. Penson (penson(AT)lptl.jussieu.fr), Mar 21, 2002

a(n) = A005493(n)-A000110(n+1). - Floor en Lyanne van Lamoen (fvlamoen(AT)hotmail.com) and Chritain Bower, Oct 16 2002. (n^2 has egf e^x * (x^2+x), a(n) thus has egf e^(e^x-1) * ( (e^x-1)^2 + (e^x-1) ) which simplifies to e^(e^x-1) * (e^2x - e^x). A005493 has egf e^(e^x+2x-1), A000110 has egf e^(e^x-1), A000110(n+1) has as egf derivative of A000110 which is e^(e^x+x-1).)

a(n) = Bell(n+2)-2*Bell(n+1). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Jul 28 2003

G.f.: sum{k>=0, k^2*x^k/prod[l=1..k, 1-lx]}. - R. Stephan, Apr 18 2004

a(n) = A123158(n,3) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 06 2006

a:=n->(sum((j-1)*stirling2(n, j), j=2..n)): seq(a(n), n=2..21); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 18 2007

Partial sums of A005494.

Sequence in context: A129164 A123347 A087439 this_sequence A048251 A017971 A017972

Adjacent sequences: A033449 A033450 A033451 this_sequence A033453 A033454 A033455

nonn

njas